Tasks 03-04 - Transformations, Composition & Inverses

Section 03: Functions as Business Models

Problem 1: Basic Transformations (x)

Given the function \(f(x) = x^2 - 2x + 3\), write the equation for each transformation:

  1. \(f(x)\) shifted up by 4 units

  2. \(f(x)\) shifted left by 3 units

  3. \(f(x)\) reflected over the x-axis

  4. \(f(x)\) stretched vertically by a factor of 2

  5. \(f(x)\) shifted right by 2 units and down by 5 units

  1. Shifted up by 4: \(g(x) = x^2 - 2x + 7\)

  2. Shifted left by 3: \(g(x) = (x+3)^2 - 2(x+3) + 3 = x^2 + 4x + 6\)

  3. Reflected over x-axis: \(g(x) = -(x^2 - 2x + 3) = -x^2 + 2x - 3\)

  4. Stretched vertically by 2: \(g(x) = 2(x^2 - 2x + 3) = 2x^2 - 4x + 6\)

  5. Right 2, down 5: \(g(x) = (x-2)^2 - 2(x-2) + 3 - 5 = x^2 - 6x + 6\)

Problem 2: Cost Function Adjustments (xx)

A manufacturing company has a cost function \(C(x) = 0.02x^2 + 15x + 2000\) where \(x\) is the number of units produced.

  • The \(0.02x^2\) term represents increasing inefficiency at high volumes
  • The \(15x\) term represents variable costs per unit (materials, labor per unit)
  • The \(2000\) represents fixed costs (rent, utilities, base salaries)

  1. Due to a new supplier contract, variable costs (both the \(0.02x^2\) and \(15x\) terms) decrease by 20%. Write the new cost function.

  2. The company moves to a larger facility, increasing fixed costs by €1,500. Write the adjusted cost function starting from the original.

  3. Combining both changes from parts (a) and (b), what is the final cost function?

  4. Calculate the cost for producing 100 units using both the original and final functions. Which is cheaper?

  5. At what production level do the original and final cost functions have the same total cost? What does this mean for the company’s decision?

  1. Variable costs decrease by 20% (multiply both variable terms by 0.8, keep fixed costs):

    • \(C_1(x) = 0.8(0.02x^2) + 0.8(15x) + 2000\)
    • \(C_1(x) = 0.016x^2 + 12x + 2000\)

  2. Fixed costs increase by €1,500 (only change the constant term):

    • \(C_2(x) = 0.02x^2 + 15x + (2000 + 1500)\)
    • \(C_2(x) = 0.02x^2 + 15x + 3500\)

  3. Combined changes (apply both transformations):

    • Start with original: \(C(x) = 0.02x^2 + 15x + 2000\)
    • Reduce variable costs by 20%: \(0.016x^2 + 12x + 2000\)
    • Increase fixed costs by €1,500: \(C_{final}(x) = 0.016x^2 + 12x + 3500\)

  4. Cost comparison at x = 100:

    • Original: \(C(100) = 0.02(100)^2 + 15(100) + 2000\)
      • \(= 200 + 1500 + 2000 = \text{€}3,700\)
    • Final: \(C_{final}(100) = 0.016(100)^2 + 12(100) + 3500\)
      • \(= 160 + 1200 + 3500 = \text{€}4,860\)
    • The final function costs €1,160 more!
    • Despite lower variable costs, the €1,500 fixed cost increase dominates at this production level.

  5. Finding the break-even production level:

    Set \(C(x) = C_{final}(x)\):

    • \(0.02x^2 + 15x + 2000 = 0.016x^2 + 12x + 3500\)
    • \(0.004x^2 + 3x - 1500 = 0\)
    • Multiply by 250: \(x^2 + 750x - 375000 = 0\)

    Using quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

    • \(x = \frac{-750 \pm \sqrt{750^2 + 4(375000)}}{2}\)
    • \(x = \frac{-750 \pm \sqrt{562500 + 1500000}}{2}\)
    • \(x = \frac{-750 \pm \sqrt{2062500}}{2}\)
    • \(x = \frac{-750 \pm 1436.3}{2}\)
    • \(x = \frac{686.3}{2} = 343.15\) units (taking positive solution)

    Interpretation:

    • Below 343 units: Original facility is cheaper (despite higher variable costs)
    • Above 343 units: New facility is cheaper (lower variable costs compensate for higher fixed costs)
    • The company should only move to the new facility if they expect to consistently produce more than 343 units!

Problem 3: Demand Curve Transformations (xx)

A product’s weekly demand is modeled by \(D(p) = 1000 - 20p\) where \(p\) is the price in euros.

  1. During a promotional campaign, demand increases by 30% at every price level. Write the new demand function.

  2. After the promotion ends, a competitor’s product launch causes the demand curve to shift, reducing demand by 200 units at every price. Write this demand function.

  3. Economic conditions change consumer behavior: due to inflation, when the actual price is \(p\) euros, consumers now perceive it as being 10% more expensive and adjust their purchasing accordingly. How does this transform the original demand function? Hint: If consumers perceive the price as 10% higher, they react as if the price were \(1.1p\) euros.

  4. Find the price that generates demand of 500 units for:

    • The original function
    • The promotional function
    • The post-competitor function
  1. 30% demand increase (vertical stretch by 1.3):
    • \(D_{promo}(p) = 1.3(1000 - 20p) = 1300 - 26p\)
  2. Demand reduced by 200 units (vertical shift down):
    • \(D_{comp}(p) = 1000 - 20p - 200 = 800 - 20p\)
  3. 10% price perception increase (horizontal transformation):
    • Consumers perceive price as \(1.1p\)
    • \(D_{inflation}(p) = 1000 - 20(1.1p) = 1000 - 22p\)
  4. Finding price for demand = 500:
    • Original: \(500 = 1000 - 20p\), so \(p = 25\) euros
    • Promotional: \(500 = 1300 - 26p\), so \(p = 30.77\) euros
    • Post-competitor: \(500 = 800 - 20p\), so \(p = 15\) euros

Problem 4: Seasonal Business Model (xxx)

An ice cream shop’s monthly profit follows the function: \[P(m) = -200(m - 7)^2 + 8000\] where \(m\) is the month (1 = January, 12 = December).

  1. In which month does the shop achieve maximum profit? What is this profit?

  2. Due to climate change, the peak season shifts 1.5 months earlier and maximum profit increases by 15%, everything else stays the same. Write the new profit function.

  3. The shop opens a second location in the Southern Hemisphere. In the Southern Hemisphere, summer occurs during Northern winter. Hence, when it’s July (peak summer) in the North, it’s January (peak winter) in the South. Write the profit function for the Southern location, assuming it has the same profit pattern from task a. but peaks in January instead of July.

  4. If both locations (North and South) operate simultaneously, write the combined monthly profit function \(P_{total}(m)\) that represents total profit across both shops based on task a.

  5. Using the combined function from part (d), find which months the total profit exceeds €10,000.

  1. Maximum profit:

    • The vertex form shows the peak at \(m = 7\) (July)
    • Maximum profit = €8,000

  2. Climate change adjustments:

    • Shift 1.5 months earlier: Change \((m - 7)\) to \((m - 5.5)\)
      • New peak: May (month 5.5)
    • 15% profit increase: Change 8000 to \(8000 \times 1.15 = 9200\)
    • New function: \(P_{climate}(m) = -200(m - 5.5)^2 + 9200\)

  3. Southern Hemisphere location:

    • Original peaks in July (month 7)
    • Southern location peaks in January (month 1)
    • Horizontal shift: Change \((m - 7)\) to \((m - 1)\)
    • \(P_{south}(m) = -200(m - 1)^2 + 8000\)

    Verification:

    • At \(m = 1\) (January): \(P_{south}(1) = -200(0)^2 + 8000 = \text{€}8,000\) ✓ (peak)
    • At \(m = 7\) (July): \(P_{south}(7) = -200(36) + 8000 = \text{€}800\) ✓ (low season)

  4. Combined profit function:

    • North: \(P_{north}(m) = -200(m - 7)^2 + 8000\)
    • South: \(P_{south}(m) = -200(m - 1)^2 + 8000\)
    • Total: Add both functions

    \[P_{total}(m) = -200(m - 7)^2 + 8000 + [-200(m - 1)^2 + 8000]\] \[P_{total}(m) = -200[(m - 7)^2 + (m - 1)^2] + 16000\]

  5. Months exceeding €10,000:

    Solve \(P_{total}(m) > 10000\):

    • \(-200[(m - 7)^2 + (m - 1)^2] + 16000 > 10000\)
    • \(-200[(m - 7)^2 + (m - 1)^2] > -6000\)
    • \((m - 7)^2 + (m - 1)^2 < 30\)

    Testing each month:

    • \(m = 1\): \((1-7)^2 + (1-1)^2 = 36 + 0 = 36 > 30\) ✗
    • \(m = 2\): \((2-7)^2 + (2-1)^2 = 25 + 1 = 26 < 30\) ✓ → \(P_{total}(2) = \text{€}10,800\)
    • \(m = 3\): \((3-7)^2 + (3-1)^2 = 16 + 4 = 20 < 30\) ✓ → \(P_{total}(3) = \text{€}12,000\)
    • \(m = 4\): \((4-7)^2 + (4-1)^2 = 9 + 9 = 18 < 30\) ✓ → \(P_{total}(4) = \text{€}12,400\)
    • \(m = 5\): \((5-7)^2 + (5-1)^2 = 4 + 16 = 20 < 30\) ✓ → \(P_{total}(5) = \text{€}12,000\)
    • \(m = 6\): \((6-7)^2 + (6-1)^2 = 1 + 25 = 26 < 30\) ✓ → \(P_{total}(6) = \text{€}10,800\)
    • \(m = 7\): \((7-7)^2 + (7-1)^2 = 0 + 36 = 36 > 30\) ✗

    Answer: Months 2, 3, 4, 5, and 6 (February through June) exceed €10,000

    Business insight: By operating in both hemispheres, the company achieves consistent high profits during spring/fall transition months when one location is entering peak season while the other is exiting it!

Problem 5: Graph Interpretation (xxx)

You are given a graph showing three functions:

  • Function A: Linear, passing through (0, 100) and (50, 600)
  • Function B: Quadratic, vertex at (30, 700), passing through origin
  • Function C: Linear, passing through (0, 400) and (40, 400)

These represent cost functions for three different production methods.

  1. Write the equation for each function.

  2. Determine the production ranges where each method is most cost-effective.

  3. A company expects to produce 35 units. Which method should they choose and what will be the cost?

  4. If fixed costs for Method A increase by €200, how does this change your recommendation for 35 units?

  1. Function equations:
    • A (linear): Slope = \(\frac{600-100}{50-0} = 10\), so \(C_A(x) = 10x + 100\)
    • B (quadratic): Vertex form with (30, 700) and passes through (0, 0)
      • \(C_B(x) = a(x - 30)^2 + 700\)
      • At origin: \(0 = 900a + 700\), so \(a = -\frac{700}{900} = -\frac{7}{9}\)
      • \(C_B(x) = -\frac{7}{9}(x - 30)^2 + 700\)
    • C (linear): Horizontal line at 400, so \(C_C(x) = 400\)
  2. Cost-effective ranges:
    • Find intersections:
      • A and C: \(10x + 100 = 400\), so \(x = 30\)
      • A and B: Solve \(10x + 100 = -\frac{7}{9}(x - 30)^2 + 700\)
      • B and C: Solve \(-\frac{7}{9}(x - 30)^2 + 700 = 400\)
    • Method C best for x < 30
    • Method A best for x > 30 (approximately)
    • Method B competitive around x = 30
  3. At 35 units:
    • Method A: \(C_A(35) = 350 + 100 = \text{€}450\)
    • Method B: \(C_B(35) = -\frac{7}{9}(25) + 700 ≈ \text{€}680\)
    • Method C: \(C_C(35) = \text{€}400\)
    • Choose Method C (€400)
  4. With A’s fixed cost increase:
    • New: \(C_A(x) = 10x + 300\)
    • At 35 units: \(C_A(35) = 350 + 300 = \text{€}650\)
    • Still choose Method C (€400)

Problem 6: Revenue Transformation Analysis (xxx)

A software company’s monthly revenue from selling \(x\) subscriptions is: \[R(x) = -2x^2 + 120x\] where \(x\) is the number of subscriptions sold (in hundreds) and revenue is in euros.

TipUnderstanding the function
  • Revenue depends on quantity sold
  • The quadratic form reflects diminishing returns (marketing costs, market saturation)
  • Each subscription currently sells for the same price

The company is analyzing different market scenarios and how they affect revenue.

  1. Find the optimal number of subscriptions to sell that maximizes revenue. What is the maximum monthly revenue?

  2. The company improves their product features. Market research shows this will increase revenue per subscription by 15% across all quantity levels (customers perceive higher value). Write the transformed revenue function \(R_{improved}(x)\) and find the new optimal quantity and maximum revenue.

  3. Due to increased competition, the company expects overall revenue to drop by 25% at every quantity level. Write the transformed revenue function \(R_{competition}(x)\) and find the new maximum revenue. How much revenue is lost compared to the original?

  4. The company considers an aggressive marketing campaign. This is expected to boost revenue by 30% at all quantity levels (better brand awareness, higher perceived value). Write the transformed revenue function \(R_{marketing}(x)\) and find its maximum revenue.

  5. Based on your calculations, rank the scenarios from best to worst in terms of maximum revenue. What do you notice about the optimal quantity in each scenario?

  1. Optimal quantity and maximum revenue (original):

    Using vertex formula: \(x = -\frac{b}{2a} = -\frac{120}{2(-2)} = \frac{120}{4} = 30\) (i.e., 3,000 subscriptions)

    Maximum revenue: \(R(30) = -2(30)^2 + 120(30) = -1800 + 3600 = \text{€}1,800\)

  2. Product improvement scenario (+15% revenue per subscription):

    This is a vertical stretch by factor 1.15 - multiply the entire function:

    • \(R_{improved}(x) = 1.15 \times (-2x^2 + 120x)\)
    • \(R_{improved}(x) = -2.3x^2 + 138x\)

    Optimal quantity: \(x = -\frac{138}{2(-2.3)} = \frac{138}{4.6} = 30\) (still 3,000 subscriptions)

    Maximum revenue: \(R_{improved}(30) = -2.3(30)^2 + 138(30) = -2070 + 4140 = \text{€}2,070\)

    Key insight: Vertical transformations don’t change the optimal x-value! The company should still aim for 3,000 subscriptions, but now earns more per subscription.

  3. Competition scenario (-25% revenue at all levels):

    This is a vertical compression by factor 0.75 - multiply the entire function:

    • \(R_{competition}(x) = 0.75 \times (-2x^2 + 120x)\)
    • \(R_{competition}(x) = -1.5x^2 + 90x\)

    Optimal quantity: \(x = -\frac{90}{2(-1.5)} = \frac{90}{3} = 30\) (still 3,000 subscriptions)

    Maximum revenue: \(R_{competition}(30) = -1.5(30)^2 + 90(30) = -1350 + 2700 = \text{€}1,350\)

    Revenue lost: €1,800 - €1,350 = €450 (25% decrease, as expected)

  4. Marketing campaign scenario (+30% revenue at all levels):

    This is a vertical stretch by factor 1.3 - multiply the entire function:

    • \(R_{marketing}(x) = 1.3 \times (-2x^2 + 120x)\)
    • \(R_{marketing}(x) = -2.6x^2 + 156x\)

    Optimal quantity: \(x = -\frac{156}{2(-2.6)} = \frac{156}{5.2} = 30\) (still 3,000 subscriptions)

    Maximum revenue: \(R_{marketing}(30) = -2.6(30)^2 + 156(30) = -2340 + 4680 = \text{€}2,340\)

  5. Ranking and key observation:

    Revenue ranking (best to worst):

    1. Marketing campaign: €2,340 (+30%)
    2. Product improvement: €2,070 (+15%)
    3. Original: €1,800 (baseline)
    4. Competition: €1,350 (-25%)

    Critical observation:

    • The optimal quantity is ALWAYS 30 (3,000 subscriptions) in every scenario!
    • This is because we applied vertical transformations (multiplying the entire function)
    • Vertical transformations scale revenue up or down but don’t change the optimal input value
    • Only the maximum revenue changes proportionally to the scaling factor

    Business insight: When external factors uniformly affect revenue across all quantity levels (better product, competition, marketing), the optimal sales target stays the same - only the profitability changes!

Problem 7: Multi-Location Profit Analysis (xxxx)

A restaurant chain has a successful location with profit function: \[P(d) = -5d^2 + 200d - 1500\] where \(d\) is the number of daily customers (in tens).

They plan to open three new locations with different market conditions:

Location A (Downtown):

  • Market research suggests profits will be 20% higher at every customer level (premium neighborhood)
  • Fixed costs increase by €500 (expensive rent)

Location B (Suburbs):

  • Profits expected to be 30% lower at every customer level (lower prices, higher costs)
  • Fixed costs reduced by €200 (cheaper rent)

Location C (Mall):

  • Peak customer time occurs 5 units earlier than the original (lunch rush vs dinner rush)
  • Profits are 10% lower at every customer level
  • Fixed costs increase by €300

  1. Write the transformed profit function for each location (A, B, and C).

  2. Find the optimal number of customers and maximum profit for each location.

  3. If the chain can only open two new locations, which combination maximizes total daily profit?

  4. For Location C specifically, explain how the horizontal shift affects when the restaurant reaches peak profit during the day.

  1. Transformed profit functions:

    Original: \(P(d) = -5d^2 + 200d - 1500\)

    Location A (Downtown):

    • 20% higher profits → multiply by 1.2
    • Fixed costs +€500 → subtract 500 more
    • Step 1: \(1.2(-5d^2 + 200d - 1500) = -6d^2 + 240d - 1800\)
    • Step 2: \(P_A(d) = -6d^2 + 240d - 1800 - 500 = -6d^2 + 240d - 2300\)

    Location B (Suburbs):

    • 30% lower profits → multiply by 0.7
    • Fixed costs -€200 → add 200
    • Step 1: \(0.7(-5d^2 + 200d - 1500) = -3.5d^2 + 140d - 1050\)
    • Step 2: \(P_B(d) = -3.5d^2 + 140d - 1050 + 200 = -3.5d^2 + 140d - 850\)

    Location C (Mall):

    • Peak 5 units earlier → replace \(d\) with \((d + 5)\) (horizontal shift left)
    • 10% lower profits → multiply by 0.9
    • Fixed costs +€300 → subtract 300 more
    • Step 1 (shift): \(P(d + 5) = -5(d+5)^2 + 200(d+5) - 1500\)
      • \(= -5(d^2 + 10d + 25) + 200d + 1000 - 1500\)
      • \(= -5d^2 - 50d - 125 + 200d + 1000 - 1500\)
      • \(= -5d^2 + 150d - 625\)
    • Step 2 (scale): \(0.9(-5d^2 + 150d - 625) = -4.5d^2 + 135d - 562.5\)
    • Step 3 (fixed costs): \(P_C(d) = -4.5d^2 + 135d - 562.5 - 300 = -4.5d^2 + 135d - 862.5\)

  2. Optimal customers and maximum profits:

    Original:

    • \(d = -\frac{200}{2(-5)} = 20\) (200 customers)
    • \(P(20) = -5(400) + 200(20) - 1500 = -2000 + 4000 - 1500 = \text{€}500\)

    Location A:

    • \(d = -\frac{240}{2(-6)} = 20\) (200 customers, same as original!)
    • \(P_A(20) = -6(400) + 240(20) - 2300 = -2400 + 4800 - 2300 = \text{€}100\)

    Location B:

    • \(d = -\frac{140}{2(-3.5)} = 20\) (200 customers, same as original!)
    • \(P_B(20) = -3.5(400) + 140(20) - 850 = -1400 + 2800 - 850 = \text{€}550\)

    Location C:

    • \(d = -\frac{135}{2(-4.5)} = 15\) (150 customers, 5 units earlier!)
    • \(P_C(15) = -4.5(225) + 135(15) - 862.5 = -1012.5 + 2025 - 862.5 = \text{€}150\)

  3. Best two-location combination:

    Daily profit at optimal levels:

    • Original: €500
    • Location A: €100
    • Location B: €550 (best!)
    • Location C: €150

    Combinations (excluding original):

    • A + B: €100 + €550 = €650
    • A + C: €100 + €150 = €250
    • B + C: €550 + €150 = €700

    Recommendation: Open Locations B and C for combined profit of €700/day

  4. Horizontal shift explanation for Location C:

    The horizontal shift left by 5 units means:

    • The original peaks at \(d = 20\) (200 customers)
    • Location C peaks at \(d = 15\) (150 customers)
    • This represents reaching peak profit 5 units (50 customers) earlier in the day
    • If customers arrive steadily throughout the day, Location C reaches maximum profitability sooner
    • This makes sense for a mall location with lunch rush vs. dinner rush

Problem 8: Function Composition Basics (x)

Given the functions:

  • \(f(x) = 3x - 2\)
  • \(g(x) = x^2 + 1\)
  • \(h(x) = \sqrt{x}\)

Find: a) \((f \circ g)(x)\) b) \((g \circ f)(x)\) c) \((f \circ g)(3)\) d) \((g \circ f)(2)\) e) \((h \circ g)(4)\)

  1. \((f \circ g)(x) = f(g(x)) = f(x^2 + 1) = 3(x^2 + 1) - 2 = 3x^2 + 1\)

  2. \((g \circ f)(x) = g(f(x)) = g(3x - 2) = (3x - 2)^2 + 1 = 9x^2 - 12x + 5\)

  3. \((f \circ g)(3) = 3(3)^2 + 1 = 27 + 1 = 28\)

  4. \((g \circ f)(2) = 9(2)^2 - 12(2) + 5 = 36 - 24 + 5 = 17\)

  5. \((h \circ g)(4) = h(g(4)) = h(17) = \sqrt{17} \approx 4.12\)

Problem 9: Finding Inverse Functions (x)

Find the inverse function for each of the following, and state the domain and range of both the original and inverse:

  1. \(f(x) = 4x - 7\)

  2. \(g(x) = \frac{x + 2}{3}\)

  3. \(h(x) = 2x^3 - 1\)

  4. \(p(x) = \frac{5}{x - 3}\) where \(x \neq 3\)

  1. \(f(x) = 4x - 7\):
    • Inverse: \(y = 4x - 7 \Rightarrow x = 4y - 7 \Rightarrow y = \frac{x + 7}{4}\)
    • \(f^{-1}(x) = \frac{x + 7}{4}\)
    • Domain/Range: Both \(f\) and \(f^{-1}\) have domain and range \(\mathbb{R}\)
  2. \(g(x) = \frac{x + 2}{3}\):
    • Inverse: \(y = \frac{x + 2}{3} \Rightarrow x = \frac{y + 2}{3} \Rightarrow y = 3x - 2\)
    • \(g^{-1}(x) = 3x - 2\)
    • Domain/Range: Both have domain and range \(\mathbb{R}\)
  3. \(h(x) = 2x^3 - 1\):
    • Inverse: \(y = 2x^3 - 1 \Rightarrow x = 2y^3 - 1 \Rightarrow y^3 = \frac{x + 1}{2}\)
    • \(h^{-1}(x) = \sqrt[3]{\frac{x + 1}{2}}\)
    • Domain/Range: Both have domain and range \(\mathbb{R}\)
  4. \(p(x) = \frac{5}{x - 3}\):
    • Inverse: \(y = \frac{5}{x - 3} \Rightarrow x = \frac{5}{y - 3} \Rightarrow y - 3 = \frac{5}{x}\)
    • \(p^{-1}(x) = \frac{5}{x} + 3\)
    • Domain of \(p\): \(\mathbb{R} \setminus \{3\}\), Range of \(p\): \(\mathbb{R} \setminus \{0\}\)
    • Domain of \(p^{-1}\): \(\mathbb{R} \setminus \{0\}\), Range of \(p^{-1}\): \(\mathbb{R} \setminus \{3\}\)

Problem 10: Production Process Composition (xx)

A manufacturing company has a three-stage production process:

Stage 1 (Raw Material Processing):

  • Cost: \(C_1(x) = 30x + 150\) where \(x\) is kg of raw material

Stage 2 (Component Manufacturing):

  • Produces: 0.7 kg of components per kg processed
  • Additional cost: \(C_2(y) = 50y + 200\) where \(y\) is kg of components

Stage 3 (Final Assembly):

  • Each kg of components yields 5 products
  • Packaging cost: \(C_3(z) = 2z + 100\) where \(z\) is number of products

  1. Express the total cost as a function of raw material quantity \(x\).

  2. How much raw material is needed to produce 100 final products?

  3. What is the total cost for producing 100 final products?

  4. Find the average cost per product when starting with 10 kg of raw material.

  1. Total cost function:
    • After Stage 1: Cost = \(30x + 150\)
    • Components produced: \(y = 0.7x\) kg
    • After Stage 2: Additional cost = \(50(0.7x) + 200 = 35x + 200\)
    • Products produced: \(z = 5(0.7x) = 3.5x\) products
    • After Stage 3: Additional cost = \(2(3.5x) + 100 = 7x + 100\)
    • Total: \(C(x) = (30x + 150) + (35x + 200) + (7x + 100) = 72x + 450\)
  2. Raw material for 100 products:
    • Need: \(3.5x = 100\)
    • \(x = \frac{100}{3.5} = 28.57\) kg
  3. Total cost for 100 products:
    • \(C(28.57) = 72(28.57) + 450 = 2057.14 + 450 = \text{€}2,507.14\)
  4. Average cost with 10 kg raw material:
    • Products: \(3.5(10) = 35\) products
    • Total cost: \(C(10) = 72(10) + 450 = \text{€}1,170\)
    • Average: \(\frac{1170}{35} = \text{€}33.43\) per product

Problem 11: Currency Exchange and Investment (xx)

An investor uses the following process:

  • Converts USD to EUR: \(E(x) = 0.85x\)
  • Invests EUR with return function: \(R(y) = 1.08y - 50\) (minimum €50 fee)
  • Converts back to USD: \(U(z) = 1.18z\)

  1. Express the final USD amount as a function of initial USD investment \(x\).

  2. What is the minimum USD investment needed to avoid losing money?

  3. Find the inverse of the overall process. What does it represent?

  4. If an investor wants to end with exactly $10,000 USD, how much should they initially invest?

  1. Final USD amount:
    • \((U \circ R \circ E)(x) = U(R(E(x)))\)
    • \(= U(R(0.85x))\)
    • \(= U(1.08(0.85x) - 50)\)
    • \(= U(0.918x - 50)\)
    • \(= 1.18(0.918x - 50)\)
    • \(= 1.08324x - 59\)
  2. Minimum to avoid loss:
    • Need: \(1.08324x - 59 \geq x\)
    • \(0.08324x \geq 59\)
    • \(x \geq \frac{59}{0.08324} = €708.89\) USD
  3. Inverse function:
    • \(y = 1.08324x - 59\)
    • \(x = 1.08324y - 59\)
    • \(y = \frac{x + 59}{1.08324}\)
    • Represents: Initial USD needed to achieve a target final amount
  4. For $10,000 final:
    • \(\frac{10000 + 59}{1.08324} = \frac{10059}{1.08324} = \text{€}9,283.48\) USD initial

Problem 12: Supply Chain Optimization (xxx)

A retailer has the following functions in their supply chain:

Wholesale Purchase:

  • Price per unit depends on quantity: \(P(x) = 100 - 0.5x\) (for \(x \leq 150\))
  • Total wholesale cost: \(W(x) = x \cdot P(x)\)

Processing & Storage:

  • Processing reduces quantity by 5% (damage/loss)
  • Storage cost: \(S(y) = 20y + 500\) where \(y\) is stored quantity

Retail Sales:

  • Can sell at: \(R(z) = 180z - z^2\) revenue for \(z\) units

  1. Express the profit as a function of wholesale purchase quantity \(x\).

  2. Find the domain restrictions for this business model.

  3. If the retailer can only store 100 units, what wholesale quantity maximizes profit?

  4. Create an inverse function that determines wholesale quantity needed for a target revenue.

  1. Profit function:
    • Wholesale cost: \(W(x) = x(100 - 0.5x) = 100x - 0.5x^2\)
    • After processing: \(y = 0.95x\) units remain
    • Storage cost: \(S(0.95x) = 20(0.95x) + 500 = 19x + 500\)
    • Revenue: \(R(0.95x) = 180(0.95x) - (0.95x)^2 = 171x - 0.9025x^2\)
    • Profit: \(\Pi(x) = 171x - 0.9025x^2 - (100x - 0.5x^2) - (19x + 500)\)
    • \(\Pi(x) = 171x - 0.9025x^2 - 100x + 0.5x^2 - 19x - 500\)
    • \(\Pi(x) = -0.4025x^2 + 52x - 500\)
  2. Domain restrictions:
    • Price constraint: \(x \leq 150\) (given)
    • Positive price: \(100 - 0.5x > 0 \Rightarrow x < 200\)
    • Practical: \(x \geq 0\)
    • Domain: \([0, 150]\)
  3. With storage limit of 100 units:
    • Need: \(0.95x \leq 100\)
    • \(x \leq 105.26\)
    • Within domain \([0, 105.26]\)
    • Vertex of profit: \(x = -\frac{52}{2(-0.4025)} = 64.6\) units
    • This is within constraint, so order 65 units
  4. Inverse for revenue target:
    • Revenue: \(R = 171x - 0.9025x^2\)
    • Solving for \(x\) given \(R\) requires quadratic formula
    • \(x = \frac{171 \pm \sqrt{171^2 - 4(0.9025)R}}{2(0.9025)}\)
    • Taking positive root: \(x = \frac{171 - \sqrt{29241 - 3.61R}}{1.805}\)

Problem 13: Market Equilibrium with Taxes (xxx)

Consider a market with:

  • Demand: \(Q_d = 1000 - 10p\)
  • Supply: \(Q_s = 5p - 50\)

The government implements a tax system:

  • Tax on consumers: \(T_c(p) = p + 5\) (perceived price)
  • Tax on producers: \(T_p(p) = 0.9p\) (received price)

  1. Find the original equilibrium without taxes.

  2. With consumer tax only, find the new equilibrium and tax revenue.

  3. With producer tax only, find the new equilibrium and tax revenue.

  4. With both taxes, express the market-clearing condition and solve for equilibrium.

  5. Find the inverse demand function and interpret its meaning with taxes.

  1. Original equilibrium:
    • \(1000 - 10p = 5p - 50\)
    • \(1050 = 15p\)
    • \(p^* = 70\), \(Q^* = 300\)
  2. With consumer tax (consumers pay \(p + 5\)):
    • Demand becomes: \(Q_d = 1000 - 10(p + 5) = 950 - 10p\)
    • New equilibrium: \(950 - 10p = 5p - 50\)
    • \(1000 = 15p\)
    • \(p = 66.67\), \(Q = 283.33\)
    • Consumers pay: €71.67
    • Tax revenue: \(5 \times 283.33 = \text{€}1,416.67\)
  3. With producer tax (producers receive \(0.9p\)):
    • Supply becomes: \(Q_s = 5(0.9p) - 50 = 4.5p - 50\)
    • New equilibrium: \(1000 - 10p = 4.5p - 50\)
    • \(1050 = 14.5p\)
    • \(p = 72.41\), \(Q = 275.9\)
    • Producers receive: €65.17
    • Tax revenue: \(0.1(72.41) \times 275.9 = \text{€}1,997.89\)
  4. With both taxes:
    • Consumers see price \(p_c = p + 5\)
    • Producers receive \(p_p = 0.9p\)
    • Demand: \(Q_d = 1000 - 10(p + 5) = 950 - 10p\)
    • Supply: \(Q_s = 5(0.9p) - 50 = 4.5p - 50\)
    • Equilibrium: \(950 - 10p = 4.5p - 50\)
    • \(1000 = 14.5p\)
    • \(p = 68.97\), \(Q = 260.3\)
  5. Inverse demand:
    • From \(Q = 1000 - 10p\): \(p = 100 - 0.1Q\)
    • With consumer tax: \(p_{actual} = 95 - 0.1Q\)
    • Represents price sellers must set to achieve quantity \(Q\) given consumer tax

Problem 14: Comprehensive Business Model (xxxx)

TechStart company has developed a complex business model:

Production:

  • Fixed costs: €10,000
  • Variable cost function: \(V(x) = 20x + 0.01x^2\)
  • Total cost: \(C(x) = 10000 + V(x)\)

Pricing Strategy:

  • Base price function: \(P(x) = 200 - 0.5x\)
  • Seasonal adjustment: \(S(p, t) = p(1 + 0.2\sin(\frac{\pi t}{6}))\) where \(t\) is month

Demand:

  • At price \(p\): \(D(p) = 2000 - 8p\)
  • Quality factor: Higher quality increases demand by factor \(Q(x) = 1 + 0.001x\)

Multi-Channel Sales:

  • Online: 60% of sales at full price
  • Retail: 40% of sales at 90% of price

  1. Express profit as a function of production quantity \(x\) (ignore seasonal effects).

  2. Find the production level where average cost equals marginal revenue.

  3. If the company wants demand of exactly 1000 units, what price should they set?

  4. During month 3 (peak season), how does the optimal price change?

  5. Create a composite function showing how production quantity affects final demand through the price mechanism.

  6. If the company can only produce between 50 and 150 units due to capacity constraints, find the optimal production level and explain whether these constraints are binding.

  1. Profit function:
    • Average price received: \(0.6P(x) + 0.4(0.9)P(x) = 0.96P(x)\)
    • Revenue: \(R(x) = x \cdot 0.96P(x) = 0.96x(200 - 0.5x) = 192x - 0.48x^2\)
    • Cost: \(C(x) = 10000 + 20x + 0.01x^2\)
    • Profit: \(\Pi(x) = 192x - 0.48x^2 - 10000 - 20x - 0.01x^2\)
    • \(\Pi(x) = -0.49x^2 + 172x - 10000\)
  2. Average cost = Marginal revenue:
    • AC: \(\frac{10000 + 20x + 0.01x^2}{x} = \frac{10000}{x} + 20 + 0.01x\)
    • MR: \(\frac{dR}{dx} = 192 - 0.96x\)
    • Set equal: \(\frac{10000}{x} + 20 + 0.01x = 192 - 0.96x\)
    • \(\frac{10000}{x} = 172 - 0.97x\)
    • \(10000 = 172x - 0.97x^2\)
    • \(0.97x^2 - 172x + 10000 = 0\)
    • Discriminant: \(172^2 - 4(0.97)(10000) = 29584 - 38800 = -9216 < 0\)
    • No real solution exists — average cost never equals marginal revenue in this range
  3. For demand = 1000:
    • \(1000 = 2000 - 8p\)
    • \(8p = 1000\)
    • \(p = \text{€}125\)
  4. Month 3 seasonal adjustment:
    • \(t = 3\): \(\sin(\frac{3\pi}{6}) = \sin(\frac{\pi}{2}) = 1\)
    • Prices increase by 20%: \(S(p, 3) = 1.2p\)
    • Optimal base price from \(\Pi(x)\): \(x = \frac{172}{0.98} = 175.5\)
    • Original price: \(P(175.5) = 200 - 87.75 = \text{€}112.25\)
    • Seasonal price: \(1.2(112.25) = \text{€}134.70\)
  5. Production → Demand composite:
    • \((D \circ P)(x) = D(P(x)) = D(200 - 0.5x)\)
    • \(= 2000 - 8(200 - 0.5x)\)
    • \(= 2000 - 1600 + 4x\)
    • \(= 400 + 4x\)
    • With quality factor: \(D_{adjusted}(x) = (400 + 4x)(1 + 0.001x)\)
  6. Optimal with constraints [50, 150]:
    • Unconstrained optimum: \(x = \frac{172}{0.98} = 175.5\)
    • This exceeds 150, so constraint is binding
    • At boundary \(x = 150\):
      • \(\Pi(150) = -0.49(22500) + 172(150) - 10000\)
      • \(= -11025 + 25800 - 10000 = \text{€}4,775\)
    • At \(x = 50\):
      • \(\Pi(50) = -0.49(2500) + 172(50) - 10000\)
      • \(= -1225 + 8600 - 10000 = -\text{€}2,625\) (loss)
    • Optimal: Produce 150 units (constraint is binding)

Problem 15: Calculator – Composite Functions (xx)

Define \(f(x) = 2x + 1\) and \(g(x) = x^2\) in your calculator.

Calculate:

  1. \(f(g(2))\)

  2. \(g(f(2))\)

  3. \(f(g(3))\)

  4. \(g(f(0))\)

With \(f(x) = 2x + 1\) and \(g(x) = x^2\):

  1. \(f(g(2))\)
    • \(g(2) = 4\)
    • \(f(4) = 2(4) + 1 = 9\)
    • Answer: 9
  2. \(g(f(2))\)
    • \(f(2) = 2(2) + 1 = 5\)
    • \(g(5) = 25\)
    • Answer: 25
  3. \(f(g(3))\)
    • \(g(3) = 9\)
    • \(f(9) = 2(9) + 1 = 19\)
    • Answer: 19
  4. \(g(f(0))\)
    • \(f(0) = 1\)
    • \(g(1) = 1\)
    • Answer: 1